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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof</dfn> The necessary condition is stated already by the previous theorem. To prove the sufficient condition, consider <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are non-zero functions. If either one of them is zero function, then they are of course linear dependent. We know</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
0=W(y_1, y_2)=\left|
\begin{array}{ll}
y_1(x) &amp; y_2(x)\\
y_1^{\prime}(x) &amp; y_2^{\prime}(x)
\end{array}
\right|=y_1(x) y_2^{\prime}(x)-y_1^{\prime}(x) y_2(x),
\end{equation*}
</div>
<p class="continuation">which implies</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\frac{y_1^{\prime}}{y_1}=\frac{y_2^{\prime}}{y_2}.
\end{equation*}
</div>
<p class="continuation">Integrating the above ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\ln |y_1|=\ln |y_2|+C~\to~y_1=C y_2~\to~y_1-C y_2=0,
\end{equation*}
</div>
<p class="continuation">where the value of <span class="process-math">\(C\)</span> depend on <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\text{.}\)</span> Anyway, it is nonzero and we can say <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are linear dependent.</p>
<span class="incontext"><a href="sec3_3.html#p-95" class="internal">in-context</a></span>
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